3.614 \(\int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=117 \[ -\frac {a \sin (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}}+\frac {\sin (c+d x)}{2 b d (a+b \cos (c+d x))^2} \]

[Out]

arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(3/2)/(a+b)^(3/2)/d+1/2*sin(d*x+c)/b/d/(a+b*cos(d*x+c
))^2-1/2*a*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))

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Rubi [A]  time = 0.13, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3022, 12, 2754, 2659, 205} \[ -\frac {a \sin (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}}+\frac {\sin (c+d x)}{2 b d (a+b \cos (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(1 - Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^3,x]

[Out]

ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]]/((a - b)^(3/2)*(a + b)^(3/2)*d) + Sin[c + d*x]/(2*b*d*(a +
b*Cos[c + d*x])^2) - (a*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 3022

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[
((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*
(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*b*(A + C)*(m + 1) - (A*b^2 + a^2*C + b^2*(A + C)*(m + 1)
)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx &=\frac {\sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac {\int \frac {\left (a^2-b^2\right ) \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=\frac {\sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac {\int \frac {\cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 b}\\ &=\frac {\sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac {a \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {b}{a+b \cos (c+d x)} \, dx}{2 b \left (a^2-b^2\right )}\\ &=\frac {\sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac {a \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {1}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )}\\ &=\frac {\sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac {a \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\operatorname {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right ) d}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2} d}+\frac {\sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac {a \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 94, normalized size = 0.80 \[ -\frac {\frac {2 \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+\frac {\sin (c+d x) (a \cos (c+d x)+b)}{(a+b \cos (c+d x))^2}}{2 d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^3,x]

[Out]

-1/2*((2*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + ((b + a*Cos[c + d*x])*Sin[c
+ d*x])/(a + b*Cos[c + d*x])^2)/((a - b)*(a + b)*d)

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fricas [A]  time = 0.50, size = 449, normalized size = 3.84 \[ \left [\frac {{\left (b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, {\left (a^{2} b - b^{3} + {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d\right )}}, \frac {{\left (b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (a^{2} b - b^{3} + {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*((b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)
*cos(d*x + c)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*
a*b*cos(d*x + c) + a^2)) - 2*(a^2*b - b^3 + (a^3 - a*b^2)*cos(d*x + c))*sin(d*x + c))/((a^4*b^2 - 2*a^2*b^4 +
b^6)*d*cos(d*x + c)^2 + 2*(a^5*b - 2*a^3*b^3 + a*b^5)*d*cos(d*x + c) + (a^6 - 2*a^4*b^2 + a^2*b^4)*d), 1/2*((b
^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*si
n(d*x + c))) - (a^2*b - b^3 + (a^3 - a*b^2)*cos(d*x + c))*sin(d*x + c))/((a^4*b^2 - 2*a^2*b^4 + b^6)*d*cos(d*x
+ c)^2 + 2*(a^5*b - 2*a^3*b^3 + a*b^5)*d*cos(d*x + c) + (a^6 - 2*a^4*b^2 + a^2*b^4)*d)]

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giac [A]  time = 0.47, size = 177, normalized size = 1.51 \[ -\frac {\frac {\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} - \frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2} {\left (a^{2} - b^{2}\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

-((pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c)
)/sqrt(a^2 - b^2)))/(a^2 - b^2)^(3/2) - (a*tan(1/2*d*x + 1/2*c)^3 - b*tan(1/2*d*x + 1/2*c)^3 - a*tan(1/2*d*x +
1/2*c) - b*tan(1/2*d*x + 1/2*c))/((a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^2*(a^2 - b^2)
))/d

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maple [A]  time = 0.07, size = 160, normalized size = 1.37 \[ \frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )^{2} \left (a +b \right )}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )^{2} \left (a -b \right )}+\frac {\arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \left (a^{2}-b^{2}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,x)

[Out]

1/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)*tan(1/2*d*x+1/2*c)^3-1/d/(a*tan(1/2*d*x+1/2*c)
^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)*tan(1/2*d*x+1/2*c)+1/d/(a^2-b^2)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x
+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 2.83, size = 148, normalized size = 1.26 \[ \frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )}{2\,\sqrt {a+b}\,\sqrt {a-b}}\right )}{d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a-b}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{a+b}}{d\,\left (2\,a\,b+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+a^2+b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(cos(c + d*x)^2 - 1)/(a + b*cos(c + d*x))^3,x)

[Out]

atan((tan(c/2 + (d*x)/2)*(2*a - 2*b))/(2*(a + b)^(1/2)*(a - b)^(1/2)))/(d*(a + b)^(3/2)*(a - b)^(3/2)) - (tan(
c/2 + (d*x)/2)/(a - b) - tan(c/2 + (d*x)/2)^3/(a + b))/(d*(2*a*b + tan(c/2 + (d*x)/2)^2*(2*a^2 - 2*b^2) + tan(
c/2 + (d*x)/2)^4*(a^2 - 2*a*b + b^2) + a^2 + b^2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)**2)/(a+b*cos(d*x+c))**3,x)

[Out]

Timed out

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